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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Migratory Birds
  5. Discussions

Migratory Birds

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  • AbhishekVermaIIT
     + 106 comments

    Since a lot of codes shared are complicated as well as sub-optimal. Here's a simple, concise and optimal approach for coders looking for an elegant solution :

    input()
count = [0]*6
for t in map(int,input().strip().split()):
    count[t] += 1
print(count.index(max(count)))

    The solution is and traverses the given list only once.

    For aspiring programmers :

    Be careful while learning from the codes you read. I haven't checked for all languages, but a lot of python-codes here, are even incorrect (including the editorial as well as currently topvoted comment). Particularly for this problem, be cautious if you see usage of "set", "dictionary" or "Counter" anywhere.

    Dictionaries/sets are unordered in python, which means one cannot trust the outcome where order matters. In fact, these solutions would always give wrong result for simple testcase like 2, 2, 1, 1 in Python 3.6. Thus, as programmers we need to ensure that our code succeeds always, and not only by chance !


    Informative Tweets for Inquisitive Minds

  • soundarkumar
     + 33 comments

    My Java 8 Solution 

    static int migratoryBirds(int n, int[] ar) {
        // Complete this function
        int ary[] = new int[n];
        for(int i = 0; i < n ; i++ )
        ary[ar[i]]++; 
        int max = 0,maxpos=0;
        for(int i = 0 ; i < n ; i++)
            {
            if(ary[i] > max)
                {
                max = ary[i];
                maxpos = i;
            }
        }
        return maxpos;
        
    }

  • nik_roby
     + 6 comments

    Using python collections FTW.

    import sys
from collections import Counter

n = int(input().strip())
types = list(map(int, input().strip().split(' ')))

birds = Counter(types)  # Counts the array into a dictionary
print(birds.most_common(1)[0][0])

  • RodneyShag
     + 8 comments

    Java solution - passes 100% of test cases

    From my HackerRank solutions.

    import java.util.Scanner;

// Time Complexity: O(n)
public class Solution {
    
    static final int NUM_TYPES = 5;
    
    static int migratoryBirds(int[] birds) {
        /* Get counts of each type */
        int[] count = new int[NUM_TYPES + 1];
        for (int num : birds) {
            count[num]++;
        }
        
        /* Find max */
        int maxIndex = 1;
        for (int i = 0; i < count.length; i++) {
            if (count[i] > count[maxIndex]) {
                maxIndex = i;
            }
        }
        return maxIndex;
    }

    public static void main(String[] args) {
        /* Save input */
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        int[] birds = new int[n];
        for (int i = 0; i < n; i++){
            birds[i] = scan.nextInt();
        }
        scan.close();
        
        /* Calculate result */
        int result = migratoryBirds(birds);
        System.out.println(result);
    }
}

  • saxtouri
     + 6 comments

    Python oneline (not optimal, though)

    print(sorted(reversed(list(set(types))), key=types.count)[-1])
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